High molecular weight paraffinic waxes in a crude oil or in a gas condensate start to precipitate when the temperature, a major parameter, decreases below the wax appearance temperature (WAT). WAT is defined the onset temperature that the solid phase wax particles start to precipitate from the continuous liquid phase. Other terms, such as cloud point, wax precipitation temperature and wax appearance point are all interchangeable. There have been various techniques used in determining the WAT values. In the case that the testing oils are transparent, ASTM D2500 can be often used to determine the WAT. Previous studies have presented the difference in WAT data when determined by three different measurement methods—microscopy, viscosity and differential scanning calorimetry. The result showed that highest WAT values were dominantly observed from microscopic measurement. The comparison of WAT values and the detection limits with various procedures is well known in the art. It should be noted that WAT is dependent upon the cooling rate and shear history. The faster cooling during pipeline transport occurs when the temperature gradient between the pipe inside and outside is significantly increased. The faster cooling condition accelerates the wax formation resulting in the higher WAT value. High mechanical shear or high flow rate, suppresses the wax formation in isothermal condition.
The pour point is another important characteristic temperature that is usually determined by ASTM D97. Flow discontinuity can occur by either wax deposition or wax gel formation. While the wax deposition can be initiated during flow, wax gel formation occurs under static conditions caused by shutdown. Subsequent to shutdown, if the wax gel develops, certain level of upstream pressure is necessary to overcome the yield stress of the gel along the pipeline for restart. Approximately two percent of precipitated wax is needed to prevent the oil flow during pour point measurement. In the static condition, the yield stress of gelled oils becomes greater during cooling, at temperatures below the pour point. Generated solid waxes in cooling environments continuously contribute to the increase of gel strength.
It is possible that oils from different origins have a same WAT value even though they contain different wax amounts and constituents. Three examples of oils in terms of the WAT data and wax constituents are listed below.
Oil with the lower WAT but faster increase of precipitated wax amount—Supposed that the WAT values of OIL-1 and OIL-2 are T1 and T2 (T1>T2), however OIL-2 shows the faster increase of wax amount during cooling. The faster increase of the precipitated wax amount in OIL-2 could be attributed to the higher contents of total wax than the total wax in OIL-1. In addition, the wax compositions are different in OIL-1 and OIL-2. Since it is well known in the art that the higher molecular weight waxes come out in an early stage from the oil phase, OIL-1 may have more wax amounts of the higher molecular weight waxes than OIL-2, however OIL-2 has more total wax content than OIL-1.
Oils with same WAT values-If there are two different oils with the same composition except with a difference in total wax compositions, the oil containing waxes of higher carbon number distribution may contain less overall wax amount than the oil containing waxes of lower carbon number distribution.
Oils with different WAT values-Oil containing more waxes can have a higher WAT as long as a wax molecular distribution is similar. More waxes of higher molecular weights also results in higher WAT.
U.S. Pat. No. 6,841,779 shows how to determine the WAT and solid wax amount versus temperature using FTIR and differential scanning calorimetry. U.S. Pat. No. 6,841,779 has several disadvantages because of its reliance on a single peak to predict the solid wax amount. The reliance on a single peak makes the solid wax amount determination less accurate because a single peak is affected by changes in the sample.
The current invention overcomes the limited access of Fourier Transform Infrared (FTIR) spectra to predict a WAT as well as a solid wax amount precipitated at temperatures below the WAT by using the reduced form derived from two peaks in the FTIR. The reduced form is not affected by variations in the sample allowing for consistent WAT determination as well as prediction of solid wax amount. This invention has the following advantages over the prior art.                The baseline shift due to variances in sample thickness does not affect the value of the reduced form in this spectral analysis        The analysis is not affected where light passes through the sample.        The analysis is not affected by sample volume.        The analysis is not affected by moisture generation on the outer surfaces of FTIR crystal windows (NaCl) during cooling process.        The analysis is not affected by a baseline shift up due to water in the oil sample        The analysis is not affected by the degree of transparency of NaCl crystal. (the NaCl crystals do not have to be transparent)        
It is important to know when and how much wax is precipitated in cooling environments where there are existing wax problems. This invention shows the enhanced applicability of FTIR spectra to determine the WAT as well as the amount of precipitated wax in temperature variance.
Analytic Background
Spectroscopic analysis follows the Beer-Lambert law shown below. There is a linear relation between a spectral absorbance and a concentration of an interest material.A=−log T=a·b·c  (Eq. 1)
A: absorbance
T: transmittance
a: absorptivity
b: pathlength
c: concentration of the material that absorbs the light source
FIG. 1 shows the FTIR spectral absorbance of the paraffin wax in molten state, in solid state, and an oil example at wavelength range of 1600-600 cm−1. The interest in wax analysis is in two regions, one is at 1402-1324 cm−1 and the other is at 735-715 cm−1. The spectral responses of these regions are well known in the art. This invention uses both wavelength regions. It should be noted that the corrected area represents the spectral integrated area subtracted the base from the total area at a designated wavelength region.
S1: corrected area at wave number range of 1402-1324 cm−1 (CH2 vibration)
S2: corrected area at wave number range of 735-715 cm−1 (CH2 rocking, significant changes in case of solid wax generation)
Supposed the resolution of FTIR is 1 cm−1.
                              S          ⁢                                          ⁢          1                =                                            ∑                              j                =                1324                            1402                        ⁢                          A              j                                =                                    (                                                                    K                    11                                    ⁢                                      C                    1                                                  +                                                      K                    12                                    ⁢                                      C                    2                                                              )                        ·            l                                              (                  Eq          .                                          ⁢          2                )                                          S          ⁢                                          ⁢          2                =                                            ∑                              j                =                715                            735                        ⁢                          A              j                                =                                    (                                                                    K                    21                                    ⁢                                      C                    1                                                  +                                                      K                    22                                    ⁢                                      C                    2                                                              )                        ·            l                                              (                  Eq          .                                          ⁢          3                )            
Both S1 and S2 were the spectral response obtained from the contribution of the liquid phase including dissolved wax and the solid phase of wax.
In (Eq. 2) and (Eq.3),
Aj: absorbance at wavelength j
l: pathlength
K: contribution factor
here, K values are classified intoK11=k1,1324+k1,1325+k1,1326+ . . . +l1,1402  (Eq.4)K12=k2,1324+k2,1325+k2,1326+ . . . +k2,1402  (Eq.5)K21=k1,715+k1,716+k1,717+ . . . +k1,735  (Eq.6)K22=k2,715+k2,716+k2,717+ . . . +k2,735  (Eq.7)
K11: Contribution factor from the solid wax at the wavelength region of 1402-1324 cm−1 
K12: Contribution factor from the liquid phase including dissolved wax at the wavelength region of 1402-1324 cm−1 
K21: Contribution factor from the solid wax at the wavelength region of 735-715 cm−1 
K22: Contribution factor from the liquid phase including dissolved wax at the wavelength region of 735-715 cm−1 
C1: Concentration of solid wax
C2: Concentration of liquid phase including dissolved wax
For example, k1,1324 represents the FTIR contribution factor of solid wax at the wavelength of 1324 cm−1, and k2,735 represents the FTIR contribution factor of liquid phase including dissolved wax at the wavelength of 735 cm−1.
By rearranging the concentration of solid wax in (Eq.2) and (Eq.3),
                              S          ⁢                                          ⁢          1                =                                            ∑                              j                =                1324                            1402                        ⁢                          A              j                                =                                    (                                                                    K                    11                                    ⁢                  x                                +                                                      K                    12                                    ⁡                                      (                                          1                      -                      x                                        )                                                              )                        ·                          l              ′                                                          (                  Eq          .                                          ⁢          8                )                                          S          ⁢                                          ⁢          2                =                                            ∑                              j                =                715                            735                        ⁢                          A              j                                =                                    (                                                                    K                    21                                    ⁢                  x                                +                                                      K                    22                                    ⁡                                      (                                          1                      -                      x                                        )                                                              )                        ·                          l              ′                                                          (                  Eq          .                                          ⁢          9                )            
Here, x and (1−x) represent the weight fraction of solid (precipitated) wax and the weight fraction of liquid phase including dissolved wax, respectively. And l′ represents the constant containing the combined contribution of pathlength and unit conversion factor when the concentration (C) in the equations (6), (7) convert to fraction (x).
As a liquid sample starts to generate the solid wax, the total volume decreases. The location FTIR light source passes is often changed during the measurement. In the case that a large number of solid wax generates, the FTIR absorbance shifts up. However, it is important that the reduced spectra area, S2/S1, does not change. It should be noted that the value change of S2 is smaller when it is compared to the value change of S1 determined at temperatures below the WAT. Solid wax affects the value of S1 in a negligible manner but increases the value of S2. The invention uses S1 as a reference value to S2 value because the sensitivity of S1 is considerably less than that of S2 in the presence of solid wax. The reduced spectral area provides a reliable value regardless of sample volume changes, existence of water, location change where FTIR light source passes. As we can see, the reduced spectral area is presented in (Eq.10).
                                                        S              ⁢                                                          ⁢              2                                            Reduced                =                              S            ⁢                                                  ⁢            2                                S            ⁢                                                  ⁢            1                                              (                  Eq          .                                          ⁢          10                )            
The reduced spectral area increases in a mildly linear manner with decreasing temperatures. The reduced spectral area data of a mineral oil shows no deviation from the linear increase because the mineral oil does not contain paraffinic waxes (data shown in Table 1).
TABLE 1FTIR data of S1, S2 and the reduced spectral area (S2/S1) froma mineral oil.Temperature(° C.)S1S2S2/S15034.58661.98410.0573663035.45462.37170.0668942035.84272.59140.0722991036.39702.88320.079215536.42422.97520.081682
However, in the case that there are paraffinic waxes in the oil, the reduced spectral area deviates significantly from the linear increase during cooling process at the temperature below the WAT. The value increase of S2 is much greater than the value change of S1. Table 2 and Table 3 present the reduced spectral area of two model oils, the 5 weight percent and the 7 weight percent of wax oils in temperature variance.
TABLE 2FTIR data of S1, S2 and the reduced spectral area (S2/S1) from5 weight percent wax model oil.Temperature(° C.)S1S2S2/S15530.43101.97860.0650195030.66142.08380.0679624530.85622.18150.0706994030.96772.27960.0736123531.16432.37020.0760553031.38252.48990.0793402531.50472.84550.0903202031.56943.21360.1017951531.65073.60190.1138021031.72553.95910.124792531.93564.26210.133459
TABLE 3FTIR data of S1, S2 and the reduced spectral area (S2/S1) from7 weight percent wax model oil.Temperature(° C.)S1S2S2/S15039.03372.65360.0679824539.33712.74520.0697874039.47822.86880.0726683539.64142.99080.0754462739.7343.58200.0901492539.87093.84160.0963512039.79054.34640.1092321540.04014.81050.1201421040.09055.27960.131692540.12455.62020.140069
As the solid waxes generate, the deviation from the mild linear increase of the reduced spectral area became greater with increasing wax amount. The difference between the measured data and the linear extrapolation of reduced spectral area implies the wax amount in the sample.
The difference of the measured data and the linear extrapolation is shown below.
                                                                                                                                                            S                        ⁢                                                                                                  ⁢                        2                                                                                    reduced                                    -                                                            S                      ⁢                                                                                          ⁢                      2                                                              S                      ⁢                                                                                          ⁢                      1                                                                                                  etp                        =                                                            S                  ⁢                                                                          ⁢                  2                                                  S                  ⁢                                                                          ⁢                  1                                            -                                                S                  ⁢                                                                          ⁢                  2                                                  S                  ⁢                                                                          ⁢                  1                                                                              etp                            (                  Eq          .                                          ⁢          11                )            
Here, subscript etp represents the extrapolation of linear increase of reduced spectral area (S2/S1).
Since there is no solid wax content in the extrapolation of reduced spectral area, K21=0
                                                                                                                                                            S                        ⁢                                                                                                  ⁢                        2                                                                    S                        ⁢                                                                                                  ⁢                        1                                                              -                                                                  S                        ⁢                                                                                                  ⁢                        2                                                                    S                        ⁢                                                                                                  ⁢                        1                                                                                                              etp                            =                            ⁢                                                                                                                  K                        21                                            ⁢                      x                                        +                                                                  K                        22                                            ⁡                                              (                                                  1                          -                          x                                                )                                                                                                                                                K                        11                                            ⁢                      x                                        +                                                                  K                        12                                            ⁡                                              (                                                  1                          -                          x                                                )                                                                                            -                                                                            K                      22                                        ⁡                                          (                                              1                        -                        x                                            )                                                                                                                          K                        11                                            ⁢                      x                                        +                                                                  K                        12                                            ⁡                                              (                                                  1                          -                          x                                                )                                                                                                                                                                    =                            ⁢                                                                    K                    21                                    ⁢                  x                                                                                            K                      11                                        ⁢                    x                                    +                                                            K                      12                                        ⁡                                          (                                              1                        -                        x                                            )                                                                                                                              (                  Eq          .                                          ⁢          12                )            
If the value of denominator is much greater than the value in numerator, the assumption that S1 considers to be a constant can be applicable (less than 5 percent of value changes, refer to Table 2 and Table 3). ThereforeK11=K12.  (Eq. 13)
Return to (Eq.12)
                                                                                          S                  ⁢                                                                          ⁢                  2                                                  S                  ⁢                                                                          ⁢                  1                                            -                                                S                  ⁢                                                                          ⁢                  2                                                  S                  ⁢                                                                          ⁢                  1                                                                          etp                =                                                            K                21                            ⁢              x                                      K              12                                =                                                    K                21                            ⁢              x                                      K              11                                                          (                  Eq          .                                          ⁢          14                )            
There are two different phases in solid wax, orthorhombic and hexagonal phases. The orthorhombic phase of solid wax transits to hexagonal phase during heating program. The left-hand-side of (Eq.14) became 0.9 for orthorhombic phase and 0.85 for hexagonal phase obtained from pure wax (x=1). The precipitated wax does not need to be orthorhombic phase. The constant was chosen to 0.85.
                                                                                                                                                                                                                                                S                              ⁢                                                                                                                          ⁢                              2                                                                                      S                              ⁢                                                                                                                          ⁢                              1                                                                                -                                                                                    S                              ⁢                                                                                                                          ⁢                              2                                                                                      S                              ⁢                                                                                                                          ⁢                              1                                                                                                                                                  etp                                        =                                                                  S                        ⁢                                                                                                  ⁢                        2                                                                    S                        ⁢                                                                                                  ⁢                        1                                                                                                              solid                            -                                                S                  ⁢                                                                          ⁢                  2                                                  S                  ⁢                                                                          ⁢                  1                                                                          molten                ≅        0.85                            (                  Eq          .                                          ⁢          15                )            
That is,
            K      21              K      11        =  0.85
(Eq.15) can be represented to
                              x          =                      1.2            ·                                          (                                                                            S                      ⁢                                                                                          ⁢                      2                                                              S                      ⁢                                                                                          ⁢                      1                                                        -                                                            S                      ⁢                                                                                          ⁢                      2                                                              S                      ⁢                                                                                          ⁢                      1                                                                                                  ept                                      )                            (                  Eq          .                                          ⁢          16                )            